Q1
Take Two numbers from user and divide both numbers but do not use the division operator.
#include <stdio.h>
S
int main(){
int a,b, div=0;
scanf("%d%d", &a, &b);
while ( (b%a) != b){
if (b >= 0){
b = b - a;
div++;
} else{
break;
}
}
printf("THe div is = %d", div);
return 0;
}
Q2
Given an array of integers of size N and an integer “d”, the task is to rotate the array elements to the left by d positions. Note: Solve the question without declaring another array. (The input array itself must be modified). Input: {1, 2, 3, 4, 5, 6, 7}, d = 2 Output: {3, 4, 5, 6, 7, 1, 2}
#include <stdio.h>
int main(){
int d, size;
scanf("%d%d", &size, &d);
int arr[size];
for (int i = 0; i < size; i++){
scanf("%d", &arr[i]);
}
for (int i = 0; i < d; i++){
int t = arr[0];
for (int j = 0; j < size-1; j++){
arr[j] = arr[j+1];
}
arr[size-1] = t;
for (int i = 0; i < size; i++){
printf("%d", arr[i]);
}
printf("\n");
}
for (int i = 0; i < size; i++){
printf("%d", arr[i]);
}
return 0;
}Q3
Take a number from user and count the sum of their digits. Input: 206 Output: 8 Input: 4569 Output: 24
#include <stdio.h>
#include <math.h>
int main(){
int n, sum = 0;
scanf("%d", &n);
int th = log10(n) + 1;
for (int i = 0; i < th; i++){
int dig = n % 10;
sum += dig;
// printf("%d", dig);
n /= 10;
}
printf("The sum is %d", sum);
return 0;
}Q4
Take a 2D array(Matrix) as input from user and check that array is symmetric or not. If array is symmetric then print “Array is Symmetric” with array if not, then print “array is not symmetric” with given array. Note: A symmetric matrix is a matrix that is equal to its transpose. Where elements in first row are equal to elements in first column and so on. Formal Definition: Aij = Aji.
#include <stdio.h>
int main(){
int r, c;
int is = 1;
scanf("%d%d", &r, &c);
int arr[r][c];
for (int i = 0; i < r; i++){
for (int j = 0; j < c; j++){
scanf("%d", &arr[i][j]);
}
}
for (int i = 0; i < r; i++){
int j;
for (j = i; j < c; j++){
if (arr[i][j] != arr[j][i]){
is = 0;
break;
}
}
if (is == 0){
break;
}
}
if (is == 1){printf("IT IS \n");}
else{printf("IT ISN'T \n");}
printf("The matrix is ");
for (int i = 0; i < r; i++){
for (int j = 0; j < c; j++){
printf("%d", arr[i][j]);
}
printf("\n");
}
}Q5
Print the following pattern on C console:
#include <stdio.h>
int main(){
int n = 5;
for (int i = 0; i < n; i++){
for(int j = 0; j < (n - i - 1); j++){
printf(" ");
}
int x = 2*i + 1;
printf("*");
if (i == 0){x = 0;}
for (int j = 0; j < (x-2); j++){
printf(" ");
}
if (i != 0){printf("*");};
printf("\n");
}
for (int i = 0; i < (n-1); i++){
for(int j = 0; j < (i+1); j++){
printf(" ");
}
printf("*");
for (int j = 0; j < (n-2*i); j++){
printf(" ");
}
if (i != (n-2)){printf("*");};
printf("\n");
}
}
Q6
Take an array with N elements as input, and then output the frequency of each element present in the array. Example: Input: {2, 4, 2, 3, 5, 5, 4, 4} Output: Frequency of 2 = 2 Frequency of 4 = 3 Frequency of 3 = 1 Frequency of 5 = 2
#include <stdio.h>
int main(){
int n, max = -100000;
scanf("%d", &n);
int arr[n];
for (int i = 0; i < n; i++){
scanf("%d", &arr[i]);
if (arr[i]>max){
max = arr[i];
}
}
int freq[max+1];
for (int i = 0; i <= (max+1); i++){
freq[i] = 0;
}
for (int i = 0; i < n; i++){
freq[arr[i]]++;
// printf("%d Should be incremented in freq by %d \n", arr[i], freq[arr[i]]);
}
for (int i = 0; i <= max; i++){
if (freq[i] > 0){
printf("Frequency of %d is %d\n", i, freq[i]);
}
}
}Q7
Take 2 2x2 matrices as input in a 2D array. Then, multiply the 2 matrices and print the final output matrix. Use the following image as a guide:
#include <stdio.h>
int main(){
int arr[2][2];
for (int i = 0; i < 2; i++){
for (int j = 0; j < 2; j++){
scanf("%d", &arr[i][j]);
}
}
int arr2[2][2];
for (int i = 0; i < 2; i++){
for (int j = 0; j < 2; j++){
scanf("%d", &arr2[i][j]);
}
}
int arr3[2][2] = {{0, 0}, {0,0}};
for (int r = 0; r < 2; r++){
for (int i = 0; i < 2; i++){
for (int j = 0; j < 2; j++){
arr3[i][j] += arr[i][r] * arr2[r][j];
}
}
}
for (int i = 0; i < 2; i++){
for (int j = 0; j < 2; j++){
printf("%d ", arr3[i][j]);
}
printf("\n");
}
}Q8
Take an array with N elements as input, and sort the array into ascending order and then print the sorted version. Input: {6, 3, 2, 7, 1, 5} Output: {1, 2, 3, 5, 6, 7}
#include <stdio.h>
int main(){
int n;
scanf("%d", &n);
int arr[n];
for (int i = 0; i < n; i++){
scanf("%d", &arr[i]);
}
for (int i = 0; i < n; i++){
printf("%d", arr[i]);
}
for (int i = 0; i < n; i++){
int max = arr[i];
int t = arr[i];
int max_pos = i;
for(int j = i; j < n; j++){
if (max < arr[j]){
max = arr[j];
max_pos = j;
}
}
arr[i] = max;
arr[max_pos] = t;
for (int i = 0; i < n; i++){
printf("\n %d", arr[i]);
}
printf("\n");
}
for (int i = 0; i < n; i++){
printf("%d", arr[i]);
}
}Q9
Junaid wants to keep track of all mobile phone bills in his X company branches. Let Y be the number of company mobile phones in each branch. Create a 2D array for bill amount, where keep track of branch ID in row subscript, mobile phone IDs in column subscript. Ask users to enter a bill for all mobile phones in all branches. Your program should print the following:
- Total bill for all branches
- Total bill for each branch
- Branch ID where maximum bill arrived
- Branch and Mobile Phone IDs where bill is highest of all mobile phones.
#include <stdio.h>
int main(){
int x, y;
int t_all = 0, b_id_max = 0, m_max = 0, bm_max;
scanf("%d%d", &x, &y);
int t_branch[x];
for (int i = 0; i < x; i++){
t_branch[i] = 0;
}
int arr[x][y];
for (int i = 0; i < x; i++){
for (int j = 0; j < y; j++){
scanf("%d", &arr[i][j]);
t_all += arr[i][j];
t_branch[i] += arr[i][j];
if (arr[i][m_max] < arr[i][j]){
m_max = j;
}
}
if (t_branch[i] > t_branch[b_id_max]){
b_id_max = i;
}
}
printf("The total bill for all %d \n", t_all);
for (int i = 0; i < x; i++){
printf("\nThe total bill for %d Branch is %d -- Maximum Mobile ID of this branch is %d", i+1, t_branch[i], m_max);
}
printf("The max branch ID is %d", b_id_max);
return 0;
}
Q10
Given an unsorted array A of size N that contains only non-negative integers, find a continuous sub-array which adds to a given number S. In case of multiple subarrays, return the subarray which comes first on moving from left to right. Input: Enter Number S = 12 Array: {1, 2, 3, 7, 5}
#include <stdio.h>
int main(){
int n, s, flag=0, loc_min=0, loc_max=0;
scanf("%d%d", &n, &s);
int arr[n];
for (int i = 0; i < n; i++){
scanf("%d", &arr[n]);
}
for(int i = 0; i < n; i++){
int j, sum = 0;
for (j = i; j < n; j++){
sum += arr[j];
printf("SUm is %d \n", sum);
if (sum == s){
flag = 1;
break;
}
}
if (flag == 1){
loc_min = i;
loc_max = j;
break;
}
}
printf("The elements from index %d to %d", loc_min, loc_max);
}Q11
You are given an n x n integer matrix grid.
Generate an integer matrix maxLocal of size (n - 2) x (n - 2) such that:
maxLocal[i][j]is equal to the largest value of the3 x 3matrix ingridcentered around rowi + 1and columnj + 1.
In other words, we want to find the largest value in every contiguous 3 x 3 matrix in grid.
Return the generated matrix.
Example 1:
Input: grid = [[9,9,8,1],[5,6,2,6],[8,2,6,4],[6,2,2,2]] Output: [[9,9],[8,6]] Explanation: The diagram above shows the original matrix and the generated matrix. Notice that each value in the generated matrix corresponds to the largest value of a contiguous 3 x 3 matrix in grid.
Example 2:
Input: grid = [[1,1,1,1,1],[1,1,1,1,1],[1,1,2,1,1],[1,1,1,1,1],[1,1,1,1,1]] Output: [[2,2,2],[2,2,2],[2,2,2]] Explanation: Notice that the 2 is contained within every contiguous 3 x 3 matrix in grid.
Constraints:
n == grid.length == grid[i].length3 <= n <= 1001 <= grid[i][j] <= 100
#include <stdio.h>
int main(){
int r = 5;
// scanf("%d", &r);
// int arr[10][10] = {{20,8,20,6,16,16,7,16,8,10},{12,15,13,10,20,9,6,18,17,6},{12,4,10,13,20,11,15,5,17,1},{7,10,14,14,16,5,1,7,3,11},{16,2,9,15,9,8,6,1,7,15},{18,15,18,8,12,17,19,7,7,8},{19,11,15,16,1,3,7,4,7,11},{11,6,5,14,12,18,3,20,14,6},{4,4,19,6,17,12,8,8,18,8},{19,15,14,11,11,13,12,6,16,19}};
// int arr[4][4] = {{9, 9, 8, 1}, {5, 6, 2, 6}, {8, 2, 6, 4}, {6, 2, 2, 2}};
int arr[5][5] = {{1,1,1,1,1}, {1,1,1,1,1}, {1,1,2,1,1}, {1,1,1,1,1}, {1,1,1,1,1}};
// int arr[3][3] = {{2,5,5},{3,2,5},{1,2,3}};
// for (int i = 0; i < r; i++){
// for (int j = 0; j < r; j++){
// scanf("%d", &arr[i][j]);
// }
// }
int res[r-2][r-2];
for (int main_r = 0; main_r < (r-2); main_r++){
for (int main_c = 0; main_c < (r-2); main_c++){
int max = -10000;
for (int sub_r = main_r; sub_r < main_r+3; sub_r++){
for (int sub_c = main_c; sub_c < main_c+3; sub_c++){
printf("Array Val %d, Pos %d, %d for the main block of %d %d\n", arr[sub_r][sub_c], sub_r, sub_c, main_r, main_c);
if (arr[sub_r][sub_c] > max){
max = arr[sub_r][sub_c];
printf("The max chosen %d\n",max);
}
}
}
res[main_r][main_c] = max;
}
}
for (int i = 0; i < (r-2); i++){
for (int j = 0; j <(r-2) ; j++){
printf("%d ", res[i][j]);
}
printf("\n");
}
}